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A particle of unit mass moves under the action of $n$ forces directed towards $n$ fixed points $A_1,A_2,...,A_n$. The force towards $A_i$ is of magnitude $k_i$ times the distance of the particle from $A_i$. When the particle is at B, it's acceleration is zero. When it is at another point $C$ which is at a distance $d$ from B, its acceleration is $f$. Find the magnitude of $f$ in terms of $k_1, k_2, ... k_n$ and $d$.

Source https://i.stack.imgur.com/4Z410.jpg

I'm interested what a proper solution to this question would be. I'm not sure if mine is correct and it feels very cheesed.

Given that the magnitude of $f$ is independent of the positions of the $A_i$ points, we can assume without loss of generality that all of the $n$ points are on top of each other. Therefore, B must also be on top of them and if I move a distance $d$ away from B, I am a distance $d$ away from all of the points. Therefore $f=d(k_1+k_2+k_3+...+k_n)$

stoic-santiago
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Mdren
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  • Something doesn't seem right here. By linearity we may replace the $n$ forces with a single force directed toward a point $A$ whose magnitude is $k$ times the distance to $A$. This $A$ is the unique point at which the force is $0$. If $B$ is at a distance $d$ from $A$, and $C$ is at a distance $d$ from $B$, then all we can say is that $C$ is at a distance $\le 2 d$ from $A$. Are you sure you have copied the question correctly? – Zhen Lin Oct 20 '20 at 13:06
  • @ZhenLin https://imgur.com/6HHMj8X – Mdren Oct 20 '20 at 13:17
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    You copied the question wrong! "When the particle is at a point B..." – TonyK Oct 20 '20 at 13:24

2 Answers2

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The first condition gives you $$ \sum_{i=1}^n (A_i - B)k_i=0 $$ and you want to compute $|\sum_{i=1}^n (A_i - C)k_i|$ but $$ \sum_{i=1}^n (A_i - C)k_i = \sum_{i=1}^n (A_i - B)k_i + \sum_{i=1}^n (B - C)k_i = \sum_{i=1}^n (B - C)k_i $$ So $$ |\sum_{i=1}^n (A_i - C)k_i| = |\sum_{i=1}^n (B - C)k_i| = |(B - C) \sum_{i=1}^n k_i| = d \sum_{i=1}^n k_i $$

Quimey
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  • Right (+1), and let's add that the force field is the same as for calculating the barycenter – G Cab Oct 20 '20 at 13:42
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Treat all points as vectors in three space. At $B$, the total force is $\sum_{i=1}^nk_i(A_i-B) = 0$ and at $C$, the total force is $\sum_{i=1}^nk_i(A_i-C) = \vec{f}$ where $\vec{f}$ is the acceleration vector of magnitude $f$ (the particle is of unit mass). Subtracting $$\sum_{i=1}^nk_i(B-C) = \vec{f}$$ giving $f = d\sum_{i=1}^nk_i$.

red whisker
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